Description
You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, ears, eyebrows and the like. A few bear no human resemblance; these typically have geometric or amorphous shapes like cubes, oil slicks or clouds of dust.
The answer is given in the 146th episode of Star Trek - The Next Generation, titled The Chase. It turns out that in the vast majority of the quadrant's life forms ended up with a large fragment of common DNA.
Given the DNA sequences of several life forms represented as strings of letters, you are to find the longest substring that is shared by more than half of them.
Input
Standard input contains several test cases. Each test case begins with 1 ≤ n ≤ 100, the number of life forms. n lines follow; each contains a string of lower case letters representing the DNA sequence of a life form. Each DNA sequence contains at least one and not more than 1000 letters. A line containing 0 follows the last test case.
Output
For each test case, output the longest string or strings shared by more than half of the life forms. If there are many, output all of them in alphabetical order. If there is no solution with at least one letter, output "?". Leave an empty line between test cases.
Sample Input
3abcdefgbcdefghcdefghi3xxxyyyzzz0
题意：求不小于 k 个字符串中的最长子串。
思路：将 n 个字符串连起来，中间用不同样的且没有出如今字符串中的字符隔开，求后缀数组。然后二分答案，将后缀分成若干组。推断每组的后缀是否出如今不小于 k 个的原串中。分两次计算。第一次二分答案，第二次输出。
#include#include #include #include #include #include typedef long long ll;using namespace std;const int maxn = 400005;int sa[maxn]; int t1[maxn], t2[maxn], c[maxn];int rank[maxn], height[maxn];void build_sa(int s[], int n, int m) { int i, j, p, *x = t1, *y = t2; for (i = 0; i < m; i++) c[i] = 0; for (i = 0; i < n; i++) c[x[i] = s[i]]++; for (i = 1; i < m; i++) c[i] += c[i-1]; for (i = n-1; i >= 0; i--) sa[--c[x[i]]] = i; for (j = 1; j <= n; j <<= 1) { p = 0; for (i = n-j; i < n; i++) y[p++] = i; for (i = 0; i < n; i++) if (sa[i] >= j) y[p++] = sa[i] - j; for (i = 0; i < m; i++) c[i] = 0; for (i = 0; i < n; i++) c[x[y[i]]]++; for (i = 1; i < m; i++) c[i] += c[i-1]; for (i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1, x[sa[0]] = 0; for (i = 1; i < n; i++) x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++; if (p >= n) break; m = p; }}void getHeight(int s[],int n) { int i, j, k = 0; for (i = 0; i <= n; i++) rank[sa[i]] = i; for (i = 0; i < n; i++) { if (k) k--; j = sa[rank[i]-1]; while (s[i+k] == s[j+k]) k++; height[rank[i]] = k; }}int k, in[maxn], r[maxn], l[maxn];char str[maxn];int check(int mid, int n, int out) { int vis[105]; int cnt = 0; memset(vis, 0, sizeof(vis)); for (int i = 1; i <= n; i++) { if (height[i] < mid) { if (out && cnt > k / 2) { for (int j = 0, m = sa[i-1]; j < mid; j++) printf("%c", r[j+m]); printf("\n"); } cnt = 0; memset(vis, 0, sizeof(vis)); } else { if (!vis[in[sa[i-1]]]) { vis[in[sa[i-1]]] = 1; cnt++; } if (!vis[in[sa[i]]]) { vis[in[sa[i]]] = 1; cnt++; } if (!out && cnt > k / 2) return 1; } } return 0;}int main() { int first = 0; while (scanf("%d", &k) != EOF && k) { int n = 0; if (first) printf("\n"); else first = 1; for (int i = 0; i < k; i++) { scanf("%s", str); l[i] = strlen(str); for (int j = n; j < n+l[i]; j++) { r[j] = str[j-n]; in[j] = i; } n += l[i] + 1; r[n-1] = 128 + i; } n--; r[n] = 0; build_sa(r, n+1, 300); getHeight(r, n); int left = 0, right = 1000, mid, ans = -1; while (left <= right) { mid = (left + right) / 2; if (check(mid, n, 0)) { ans = mid; left = mid + 1; } else right = mid - 1; } if (ans <= 0) printf("?
\n"
); else check(ans, n, 1); } return 0; }